AP EAMCET · PHYSICS · Center of Mass Momentum and Collision
A ball falls freely from a height of \(180 \mathrm{~m}\) on to a hard horizontal floor and repeatedly bounces. If the coefficient of restitution is 0.5 , the average speed and average velocity of the ball before it ceases to rebound are respectively
(acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )
- A \(10 \mathrm{~ms}^{-1}, 10 \mathrm{~ms}^{-1}\)
- B \(50 \mathrm{~ms}^{-1}, \frac{50}{3} \mathrm{~ms}^{-1}\)
- C \(\frac{50}{3} \mathrm{~ms}^{-1}, 10 \mathrm{~ms}^{-1}\)
- D \(\frac{20}{3} \mathrm{~ms}^{-1}, \frac{50}{3} \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(\frac{50}{3} \mathrm{~ms}^{-1}, 10 \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
When ball dropped from height \(h\), then time taken to reach the ground \( t_0=\sqrt{\frac{2 h}{g}} \text { and speed, } v_0=\sqrt{2 g h} \) After first collision, its speed will become \(v_1=e v_0=e \sqrt{2 g h}\) where, \(e=\) coefficient of restitution. Now, the ball will go…
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